# Tag Info

## New answers tagged collision

0

I would think the second situation causes more damage. The total energy of the first situation is $$2\cdot\frac 12 m v^2$$ The second situation total energy is $$\frac 12 m (2v)^2=4\cdot\frac 12 m v^2$$ After the collision, car plastic deformation energy can be approximated by the total kinetic energy. Well you can argue there are other way to consume the ...

0

@Numrok 's analysis makes certain assumptions which are then mentioned in the last paragraph of the answer. For simplicity assume that the speed of the cars is measured in m/s rather than km/hr. In the head on collision with both cars of the same mass and speed the final kinetic energy will be zero if the cars interlock as a result of the collision. ...

0

in this answer im not substituting values with proper units as it is all just about comparrison there are two ways of refering impact: force and impulse and for each body impulse is different impulse on an body = change in its momentum ultimately if you measure impact as impulse, first case both car have same change in velocity (30 to 0) if masses are ...

0

I think you should look at Newton's 3rd law in a frame where forces are balanced and the initial impact transition has settled. e.g. when you hit the wall with 50 lbs force, even if you have good muscle control and apply close to 50 lbs to your hand, it does not move in a linear acceleration because it has to fight its way through a complex multi degree of ...

51

You've caught a non-intuitive part of Newton's 3rd law. It's actually applying in the case you mention, but because the objects involved are of dissimilar hardness it's easy to perceive the impact as a violation of the law. Impacts are actually really complicated. Consider this slow motion video of a punch to the gut. We won't be able to cover all of the ...

15

What makes you think that the maximum force you applied to the dry wall was anything like the maximum force you applied to the brick? It certainly wasn't. The dry wall gave way much before you were able to attain the same force as applied to the brick. Try punching the air and see how much force you are able to apply. The experimental evidence that the ...

7

TL;DR: The physics of hitting things are not as easy as exerting a constant force on something. What I am trying to say with that is that Newton's law of course applies, but it would be more obvious to see it if you were just pushing/leaning against the wall with your weight. Then I'd say the two walls probably feel roughly the same. So what is different ...

11

There is no doubt the Newton's third law holds in this case. The source of confusion is the fact that you are neglecting the time interval of the collision as well as the momentum change the colliding body. As we shall see it is incorrect to assume you applied the same force in both cases just because you started with the same initial conditions, i.e. the ...

3

The more fundamental thing to understand are the conservation laws, particularly the conservation of momentum and of energy. The availability of energy gives rise to force. When you move your fist towards an object at a particular velocity you contain within it a kinetic energy and momentum. Materials are held together with binding forces that, at the ...

5

Car collision "damage" usually goes with the energy in the zero momentum frame. In both cases that is (since in the zero momentum frame, the two cases are equivalent, assuming the masses of the cars are equal): $$E_1 = 2 \times \frac{1}{2} m v_{rel}^2 = m \left( 30 \frac{km}{h} \right)^2$$ Therefore a priori there is no difference between the two situations....

4

It can't fall slower as the first cosmical speed (7.8 km/s), which is still very high. Although it would cause much smaller destruction as it would hit directly with the mean speed of the meteors (10-70km/s). The lower angle of the hit doesn't play a significant role, because considering its mass, the interaction with the atmosphere will be probably ...

0

Assume all masses are the same and we have a perfectly symmetrical collision. It seems a logical assumption doesn't it? Then, try to understand the following conservation of energy equation. Use it to solve for $v_{af}$: $$\frac{1}{2}mv_{ai}^2 = \frac{1}{2}mv_{af}^2 + \frac{1}{2}mv_{b}^2 +\frac{1}{2}mv_{c}^2$$ Once you've done that, go back and solve the ...

0

You will agree that the velocity of A depends on the elasticity of the collision. Now, momentum is conserved regardless of the elasticity of the collision. Therefore, you cannot expect momentum conservation alone to determine the outgoing velocity of A. In detail, the total outgoing momentum depends on two unknowns, i.e. A's velocity and B and C's common ...

1

This is an interesting question and one that probably needs detailed simulation to settle. But one can make the following broad prediction: the shape of the meteorite would have minimal effect on the outcome, for the following reasons: At the kinds energies let slip in the moments of impact and the kinds of pressures and temperatures that prevail, all ...

4

If I understand correctly, you are asking if a meteor impact could (i) slow the Earth's rotation on its axis or revolution around the Sun enough to account for the 8 to 12-fold decrease in longevity of human-kind measured in Earth days/years; and (ii) cause 40 days of torrential rain, resulting in sufficient inland flooding to float a large wooden boat. An ...

0

Assuming point particles and Newtonian gravity they will collide iff, in an inertial frame based in the centre of mass of the system: angular momentum is zero; total energy measured is less than zero. Where the zero of potential energy is when the particles are infinitely far apart. (1) means that the velocities are radial, (2) means there is a time ...

0

If the particles collide, it means they couldn't escape each other given their initial conditions of mass, locations, and velocities. For example, two masses of any value, positioned at any location, will definitely collide if their initial velocities are zero with regard to some reference grid you've pre-selected. Their initial kinetic energies, zero in ...

1

For inelastic scattering, the initial momentum is $m_b v_{b_i}$. After collision, both $m_b$ and $m_c$ move together, with a velocity $v_{b_f}=v_{c_f}=v_{cm}$. By conservation of momentum $m_b v_{b_i}=m_b v_{b_f}+m_c v_{c_f}=(m_b +m_c)v_{cm}$, whichyield the equation that you are looking for

0

Definitions Let us first start with some definitions of parameters, in no particular order. I will be describing elastic collisions, assuming a quasi-neutral (i.e., $n_{e} = \sum_{s} \ n_{s} \ Z_{s}$) plasma. Thus, the collisions involve long-range forces and are called Coulomb collisions. Constants $e$ = the elementary charge $\varepsilon_{o}$ = the ...

1

When the man throws the ball, both the ball and the man get equal momentum in the opposite directions. Since the collusion is elastic, i.e: no loss in energy, the ball rebounds with momentum of the same magnitude but in the opposite direction. At this point, both the ball and the man have momentum in the same direction with equal mangitude. When the man ...

0

Straight after the man throws the ball, his velocity is determined by conservation of linear momentum, that is, the momentum of the man recoiling is equal to the momentum of the ball leaving his hand: $Mv_1=mv$ After the ball bounces elastically off the wall, it returns towards the man at opposite velocity (equal in magnitude, opposite direction), so ...

1

When a man in frictionless surface throws the ball in forward direction, by conservation of linear momentum he gets pushed back (exactly the case in space where astronaut throws something back to move foreward).Here,when man throws the ball, the momentum of ball and man are exactly equal and their velocities are in opposite direction. But you need to note ...

0

In Elastic collisions the interaction forces are conservative. We can represent the total Energy of the System as : E = U + K When the particles are far away from each other (separation > 2R) their potential energies remain constant which I choose to be U. This is true except when the particles are in contact which other. After collision the colliding ...

1

Let me switch to vector notation where $v_i$ is the velocity (vector) of ball $i$ prior to collision, and $v^\prime_i$ is the velocity (vector) immediately after collision. When the collision occurs, the two balls impart an equal but opposite impulse on one another of magnitude $J$. The velocities post-collision are then given by \begin{align} v_1^\prime&...

0

If you impose a non slip condition, then you say that during the collision both disks have the same tangential speed. You can calculate this speed (both the translational speed and the rotational speed contribute) and impose a final common tangential speed, plus the conservation of the angular momentum. The problem with this is that now you have an ...

1

Because your "balls box" runs into problems with propagation delays, etc, I decided to rearrange the problem a little bit - then I think it becomes solvable. In the scenario you want to describe, you have three balls: one with mass $m_0$ moving with velocity $v$ to the right; a second with mass $m_0$ moving with velocity $v$ to the left; and a third with ...

1

When an asteroid collides with the Earth at a speed of, say, 30 km/s, the asteroid will totally disintegrate. The energy involved in the collision between an oxygen atom in the asteroid and an oxygen atom in the Earth is about 75 eV, which is way more than the binding energy. So, the chemical bonds between the atoms in the matter directly involved in the ...

1

Impulse is a vector ($\overrightarrow J= \overrightarrow F \triangle t$) and its absolute value is J. Thus it has two components in this case and the vector can be written $(J_x,J_y)$. where, $$J_x=J\cdot \cos(\theta)=J\cdot \frac{\triangle rx_i + \triangle rx_j}{\sigma_i+sigma_j}$$ And $J_y=J\cdot \sin(\theta)$ can be obtained similarly.

1

You have proved with your analysis that $v_2$ cannot be zero. $m_2$ must be moving with some non-zero velocity, either in the same direction as $m_1$, in which case $v_2$ must be smaller than $v_1$ (or $m_1$ will never catch up to $m_2$) or $m_2$ must be moving in the opposite direction to $m_1$. With the information given, there are 4 unknowns: \$v_1, v_2, ...

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